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Fourier Series - Square Wave

  1. Oct 1, 2015 #1
    Hello,

    I think that I have done this correctly, but this is the first problem I have done on my own and would appreciate confirmation.

    1. The problem statement, all variables and given/known data


    Find the Fourier series corresponding to the following functions that are periodic over the interval [itex](−π, π)[/itex] with: (a) [itex]f(x) = 1[/itex] for [itex]−\frac{π}{2} < x < \frac{π}{2}[/itex] and [itex]f(x) = 0[/itex] otherwise.

    2. Relevant equations


    3. The attempt at a solution

    The first coefficient [itex]a_{0} = \frac{1}{\pi} \int^{0.5\pi}_{-0.5\pi} f(x) dx = 1[/itex]

    [itex]a_{n} = \frac{1}{\pi} \int^{0.5\pi}_{-0.5\pi} \cos(nx) dx[/itex]

    which leads to the following;

    [itex]a_{n} = - \frac{1}{n \pi} ((-1^{n}) - 1)[/itex]

    [itex]b_{n} = \frac{1}{\pi} \int^{0.5 \pi}_{-0.5 \pi} \sin(nx) dx = 0[/itex]

    so my Fourier series is;

    [itex]f(x) = 1/2 + \frac{2}{\pi}(\cos(x) - \frac{1}{3}\cos(3x) + \frac{1}{5}\cos(5x) - ... + ) [/itex]

    or [itex]f(x) = \frac{1}{2} + \Sigma^{\infty}_{n = 1} (- \frac{1}{n \pi}( (-1)^{n} - 1)) \cos(nx)[/itex]

    Does this look ok?

    Thanks
     
    Last edited: Oct 1, 2015
  2. jcsd
  3. Oct 1, 2015 #2
    all sins looks good. seems to add up to 1 at origin.
     
  4. Oct 1, 2015 #3
    I think I have made a mistake somewhere, based on some plots I made in mathematica.

    It does not appear to be creating a square wave.

    I will go through it again.
     
  5. Oct 1, 2015 #4
    Aha! I have found the mistake.

    The [itex]a_{n}[/itex] coefficients are related to [itex]\cos(nx)[/itex] not [itex]\sin(nx)[/itex]
     
  6. Oct 1, 2015 #5

    Ray Vickson

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    No, it is not OK. Your formula for the coefficients give results that are either 0 or positive, because ##-[(-1)^n - 1] = 1 - (-1)^n## is 0 for even n and 2 for odd n. In fact, the ##a_n## for successive odd n should alternate in sign. (However, when you subsequently wrote out the first few terms of the series, you did have alternating signs. How did that happen?)
     
  7. Oct 1, 2015 #6
    I computed several coefficients for a, which lead me to observe the alternating sign pattern. My mistaken formula was not used there, I evaluated the expression for each.

    I don't know if this is the prettiest formula, but it does the trick.

    [itex]a_{n} = \frac{2}{\pi + 2(n-1) \pi} (-1)^{n-1}[/itex]
     
  8. Oct 1, 2015 #7

    Ray Vickson

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    Another one is
    [tex] a_n = \frac{2}{\pi} \frac{\sin(n \pi/2)}{n}. [/tex]
     
  9. Oct 1, 2015 #8
    This result pops out very easily when performing the integral that defines a_n.

    I was trying to find an algebraic expression to model that, mainly because it's what I saw on an example in class. Am I just making things harder on myself, or is there any benefit to showing it like this?
     
  10. Oct 1, 2015 #9

    Ray Vickson

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    Well, the numerator of the second fraction is 0 for even n, is +1 for n = 1, 5, 9, ... and is -1 for n = 3, 7, 11, ... .All you need to do is find a nifty way of describing that in a formula that does not involve trig functions, etc. It might be easier to write the relevant n in the form n = 2m+1, m = 0, 1, 2, ... . Now we need +1 for even m and -1 for odd m, so ##(-1)^m## will do it. That is,
    [tex] a_{2m+1} = \frac{2 (-1)^m}{\pi (2m+1)} [/tex]
     
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